Polynomial roots

I've just noticed that every polynomial of degree , with , has to at least have one real root (because complex numbers happen in pairs).
This is true, right? It's shown in my A-level textbook that if is a root of a polynomial equation with real coefficients, then is
also a root of the polynomial equation. But I doubt showing just that would be sufficient. How can it be shown?

I've just noticed that every polynomial of degree , with , has to at least have one real root (because complex numbers happen in pairs).
This is true, right? It's shown in my A-level textbook that if is a root of a polynomial equation with real coefficients, then is
also a root of the polynomial equation. But I doubt showing just that would be sufficient. How can it be shown?

2k+1 is odd; therefore, [iMATH]x^{2k+1}[/iMATH] must have at least one real solution. Actually induction may work here.

For roots of the polynomial, we only have 2 options either or . If , then .

But your conclusion is exactly what I'm trying to prove. I've noticed it while sketching odd
degree polynomials, but I have written it in that way (because I thought it might be better).
And thanks, I don't know I was thinking when I put the restriction that .

I've just noticed that every polynomial of degree , with , has to at least have one real root (because complex numbers happen in pairs).
This is true, right? It's shown in my A-level textbook that if is a root of a polynomial equation with real coefficients, then is
also a root of the polynomial equation. But I doubt showing just that would be sufficient. How can it be shown?

Hello friend,

Your hypothesis is true. Namely that if then for some

There are many proofs of this. One, as you mentioned is that by the fundamental theorem of algebra has exactly zeros (counting multiplicity), then we know that . Now, it is fairly easy to show that if all of the betas are non-real that must have an imaginary coefficient contradictory to our hypothesis. Thus, must have a real zero.

Thinking more along the lines of analysis is continuous. But, clearly (why?) and thus at some point and at some point and the intermediate value theorem implies the existence of a zero.

I've just noticed that every polynomial of degree , with , has to at least have one real root (because complex numbers happen in pairs).
This is true, right? It's shown in my A-level textbook that if is a root of a polynomial equation with real coefficients, then is
also a root of the polynomial equation. But I doubt showing just that would be sufficient. How can it be shown?